zoukankan      html  css  js  c++  java
  • poj2109 Power of Cryptography

    Description

    Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be only of theoretical interest. 
    This problem involves the efficient computation of integer roots of numbers. 
    Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the nth. power, for an integer k (this integer is what your program must find).

    Input

    The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10101 and there exists an integer k, 1<=k<=109 such that kn = p.

    Output

    For each integer pair n and p the value k should be printed, i.e., the number k such that k n =p.

    Sample Input

    2 16
    3 27
    7 4357186184021382204544

    Sample Output

    4
    3
    1234
    这题可以用double型做,

    类型            长度 (bit)           有效数字                   绝对值范围

        float            32                     6~7                 10^(-37) ~ 10^38

        double         64                    15~16              10^(-307) ~10^308

        long double  128                  18~19               10^(-4931) ~ 10 ^ 4932

    因为k^n=p,所以k=p^(1.0/n);

    #include<stdio.h> #include<math.h> int main() { double n,m; while(scanf("%lf%lf",&n,&m)!=EOF)     {     printf("%.0f ",pow(m,(1.0)/n));     }     return 0; }

    也可以用二分法做

    #include<stdio.h> #include<string.h> #include<math.h> #define eps 1e-8 int main() { double n,m,i,j,h,temp; int l,r,mid; while(scanf("%lf%lf",&n,&m)!=EOF) { r=1000000000; l=1; while(l<=r){ mid=(l+r)/2; temp=pow(mid,n); //printf("%lf       %lf ",temp,pow(mid,n)); if(fabs(pow(mid,n)-m)<eps){ printf("%d ",mid);break; //因为这里求的是一个确定的整数,所以可以直接输出 } else if(pow(mid,n)<m){ l=mid+1; } else if(pow(mid,n)>m){ r=mid-1; } } } return 0; }

  • 相关阅读:
    EF初始化mysql数据库codefirst
    css盒子模型、文档流、相对与绝对定位、浮动与清除模型
    微信群打卡机器人XiaoV项目开源 | 蔡培培的独立博客
    关于12306Bypass-分流抢票
    剑指Offer刷题总结
    写在前面
    mysql/mongo/nginx手册整理(2021版)
    linux(centos7) 查看磁盘空间大小
    vue中axios.post的复杂参数传参不支持的解决办法
    webApi跨域Cross问题的简单解决
  • 原文地址:https://www.cnblogs.com/herumw/p/9464824.html
Copyright © 2011-2022 走看看